Question

introduction to derivatives: problem 1 (1 point) results for this submission 5 of the answers are not correct. let ( f(x) = x^3 - 15x ). calculate the difference quotient ( \frac{f(3+h) - f(3)}{h} ), for ( h = ) box, ( h = .01 ) box, ( h = -.01 ) box, ( h = -.1 ) box. if someone now told you that the derivative (slope of the tangent line to the graph) of ( f(x) ) at ( x = 3 ) was an integer, what would you expect it to be? box note: you can earn partial credit on this problem. preview my answers submit answers your score was recorded. your score was successfully sent to canvas. you have attempted this problem 2 times. you received a score of 0% for this attempt. your overall recorded score is 0%. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Find \( f(3 + h) \) and \( f(3) \)

Given \( f(x)=x^{3}-15x \), first calculate \( f(3 + h) \):
\( f(3 + h)=(3 + h)^{3}-15(3 + h) \)
Expand \( (3 + h)^{3} \) using the formula \( (a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3} \), where \( a = 3 \) and \( b = h \):
\( (3 + h)^{3}=3^{3}+3\times3^{2}\times h+3\times3\times h^{2}+h^{3}=27 + 27h+9h^{2}+h^{3} \)
Then \( f(3 + h)=27 + 27h+9h^{2}+h^{3}-45 - 15h=h^{3}+9h^{2}+12h - 18 \)
Next, calculate \( f(3) \):
\( f(3)=3^{3}-15\times3=27 - 45=-18 \)

Step2: Calculate the difference quotient \( \frac{f(3 + h)-f(3)}{h} \)

Substitute \( f(3 + h) \) and \( f(3) \) into the difference quotient:
\( \frac{f(3 + h)-f(3)}{h}=\frac{(h^{3}+9h^{2}+12h - 18)-(-18)}{h}=\frac{h^{3}+9h^{2}+12h}{h} \)
Simplify by dividing each term in the numerator by \( h \) (assuming \( h
eq0 \)):
\( \frac{h^{3}+9h^{2}+12h}{h}=h^{2}+9h + 12 \)

Step3: Calculate for different \( h \) values

• For \( h = 1 \):

Substitute \( h = 1 \) into \( h^{2}+9h + 12 \):
\( 1^{2}+9\times1 + 12=1 + 9+12 = 22 \)

• For \( h = 0.1 \):

Substitute \( h = 0.1 \) into \( h^{2}+9h + 12 \):
\( (0.1)^{2}+9\times0.1 + 12=0.01+0.9 + 12 = 12.91 \)

• For \( h=- 0.1 \):

Substitute \( h=-0.1 \) into \( h^{2}+9h + 12 \):
\( (-0.1)^{2}+9\times(-0.1)+12=0.01 - 0.9+12 = 11.11 \)

• For \( h=-1 \):

Substitute \( h = - 1 \) into \( h^{2}+9h + 12 \):
\( (-1)^{2}+9\times(-1)+12=1 - 9 + 12 = 4 \)

Step4: Find the derivative at \( x = 3 \)

The derivative of \( f(x) \) at \( x = 3 \) is the limit of the difference quotient as \( h\to0 \). We can also recall that for \( f(x)=x^{3}-15x \), the derivative \( f^\prime(x)=3x^{2}-15 \). Substitute \( x = 3 \):
\( f^\prime(3)=3\times3^{2}-15=27 - 15 = 12 \)

For \( h = 1 \):

Step1: Use the simplified difference quotient

We have the difference quotient \( \frac{f(3 + h)-f(3)}{h}=h^{2}+9h + 12 \).

Step2: Substitute \( h = 1 \)

Substitute \( h = 1 \) into \( h^{2}+9h + 12 \):
\( 1^{2}+9\times1+12 = 1 + 9 + 12=22 \)

Step1: Use the simplified difference quotient

We have the difference quotient \( \frac{f(3 + h)-f(3)}{h}=h^{2}+9h + 12 \).

Step2: Substitute \( h = 0.1 \)

Substitute \( h = 0.1 \) into \( h^{2}+9h + 12 \):
\( (0.1)^{2}+9\times0.1+12=0.01 + 0.9+12 = 12.91 \)

Step1: Use the simplified difference quotient

We have the difference quotient \( \frac{f(3 + h)-f(3)}{h}=h^{2}+9h + 12 \).

Step2: Substitute \( h=-0.1 \)

Substitute \( h = - 0.1 \) into \( h^{2}+9h + 12 \):
\( (-0.1)^{2}+9\times(-0.1)+12=0.01-0.9 + 12=11.11 \)

Answer:

\( 22 \)

For \( h = 0.1 \):